Question

m = 1.75 kg is released from rest at a height h = 60.0 cm above a light vertical spring of forceconstant k as in Figure [a] shown below. The ball strikes the top of the spring and compresses it a distance d = 8.90 cmas in Figure [b] shown below. Neglecting any energy losses during the collision, find the following.

(a) Find the speed of the ball just as it touches the spring.
m/s

(b) Find the force constant of the spring.
N/m

Answer 1

given thatmass of ball m = 1.75 kgcompressed distance x = 0.0890 m

a) speed of the spring v = √2gh= √2 ( 9.8 m/s²)(0.6 m)= 3.43 m/s

b)spring constant k = mg/x=( 1.75 kg)( 9.8 m/s²)/0.0890 m= 192.69 N/m

Answer 2

a) u=0
s=0.6m
a=g
v^2 = u^2 +2as = 2as = 2*9.8*0.6
v= 3.43m/s
b) d= 8.9cm
momentum per second (mv/t) =force =kx
f=kx = mv/t
per second means t=1
so k= mv/x = 1.75*3.43/.089 = 67.44 N/m